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Question Number 15570 by tawa tawa last updated on 11/Jun/17

Solve: 2^x = x^4

$$\mathrm{Solve}:\:\:\:\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{x}^{\mathrm{4}} \\ $$

Commented by tawa tawa last updated on 11/Jun/17

workings

$$\mathrm{workings} \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17

Commented by mrW1 last updated on 12/Jun/17

the graph is much bigger than what you display. there is a third zero point on the right side.

$$\mathrm{the}\:\mathrm{graph}\:\mathrm{is}\:\mathrm{much}\:\mathrm{bigger}\:\mathrm{than}\:\mathrm{what} \\ $$$$\mathrm{you}\:\mathrm{display}.\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{third}\:\mathrm{zero} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{right}\:\mathrm{side}. \\ $$

Commented by tawa tawa last updated on 12/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$ \\ $$

Answered by mrW1 last updated on 12/Jun/17

2^x =x^4 2^(x/4) =±x e^((x/4)ln 2) =±x ±1=xe^(−((xln 2)/4)) ∓((ln 2)/4)=(−((xln 2)/4))e^((−((xln 2)/4))) ⇒−((xln 2)/4)=W(∓((ln 2)/4)) x=((W(∓((ln 2)/4)))/(−((ln 2)/4)))=((W(∓0.173287))/(−0.173287)) = { ((((−2.772589)/(−0.173287))=16)),((((−0.214811)/(−0.173287))=1.2396)),((((0.149260)/(−0.173287))=−0.8613)) :}

$$\mathrm{2}^{\mathrm{x}} =\mathrm{x}^{\mathrm{4}} \\ $$$$\mathrm{2}^{\frac{\mathrm{x}}{\mathrm{4}}} =\pm\mathrm{x} \\ $$$$\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}} =\pm\mathrm{x} \\ $$$$\pm\mathrm{1}=\mathrm{xe}^{−\frac{\mathrm{xln}\:\mathrm{2}}{\mathrm{4}}} \\ $$$$\mp\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}=\left(−\frac{\mathrm{xln}\:\mathrm{2}}{\mathrm{4}}\right)\mathrm{e}^{\left(−\frac{\mathrm{xln}\:\mathrm{2}}{\mathrm{4}}\right)} \\ $$$$\Rightarrow−\frac{\mathrm{xln}\:\mathrm{2}}{\mathrm{4}}=\mathrm{W}\left(\mp\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}\right) \\ $$$$\mathrm{x}=\frac{\mathrm{W}\left(\mp\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}\right)}{−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}}=\frac{\mathrm{W}\left(\mp\mathrm{0}.\mathrm{173287}\right)}{−\mathrm{0}.\mathrm{173287}} \\ $$$$=\begin{cases}{\frac{−\mathrm{2}.\mathrm{772589}}{−\mathrm{0}.\mathrm{173287}}=\mathrm{16}}\\{\frac{−\mathrm{0}.\mathrm{214811}}{−\mathrm{0}.\mathrm{173287}}=\mathrm{1}.\mathrm{2396}}\\{\frac{\mathrm{0}.\mathrm{149260}}{−\mathrm{0}.\mathrm{173287}}=−\mathrm{0}.\mathrm{8613}}\end{cases} \\ $$

Commented by tawa tawa last updated on 12/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$ \\ $$

Commented by Tinkutara last updated on 12/Jun/17

Sir, how do you calculate W(± 0.173287)? Is there any method to calculate this without calculator? How to find this using Geogebra?

$$\mathrm{Sir},\:\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{calculate}\:{W}\left(\pm\:\mathrm{0}.\mathrm{173287}\right)? \\ $$$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{method}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{this} \\ $$$$\mathrm{without}\:\mathrm{calculator}?\:\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\mathrm{this} \\ $$$$\mathrm{using}\:\mathrm{Geogebra}? \\ $$

Commented by mrW1 last updated on 12/Jun/17

Normal calculators don′t have this Lambert function. There are some online calculators for W−function, but most of them give only one value for x<0. With geogebra you can calculate W−function value by yourself: For example you want to calculate W(−((ln 2)/4)). That is the root(s) of f(x)=xe^x +((ln 2)/4).

$$\mathrm{Normal}\:\mathrm{calculators}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{this} \\ $$$$\mathrm{Lambert}\:\mathrm{function}.\:\mathrm{There}\:\mathrm{are}\:\mathrm{some} \\ $$$$\mathrm{online}\:\mathrm{calculators}\:\mathrm{for}\:\mathrm{W}−\mathrm{function}, \\ $$$$\mathrm{but}\:\mathrm{most}\:\mathrm{of}\:\mathrm{them}\:\mathrm{give}\:\mathrm{only}\:\mathrm{one}\:\mathrm{value} \\ $$$$\mathrm{for}\:\mathrm{x}<\mathrm{0}. \\ $$$$\mathrm{With}\:\mathrm{geogebra}\:\mathrm{you}\:\mathrm{can}\:\mathrm{calculate} \\ $$$$\mathrm{W}−\mathrm{function}\:\mathrm{value}\:\mathrm{by}\:\mathrm{yourself}: \\ $$$$\mathrm{For}\:\mathrm{example}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{calculate} \\ $$$$\mathrm{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}\right).\:\mathrm{That}\:\mathrm{is}\:\mathrm{the}\:\mathrm{root}\left(\mathrm{s}\right)\:\mathrm{of} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{xe}^{\mathrm{x}} +\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}. \\ $$

Commented by mrW1 last updated on 12/Jun/17

Commented by Tinkutara last updated on 12/Jun/17

Do you mean that in general if we want W(n), these are the roots of f(x) = xe^x − n ?

$$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{that}\:\mathrm{in}\:\mathrm{general}\:\mathrm{if}\:\mathrm{we} \\ $$$$\mathrm{want}\:{W}\left({n}\right),\:\mathrm{these}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${f}\left({x}\right)\:=\:{xe}^{{x}} \:−\:{n}\:? \\ $$

Commented by mrW1 last updated on 12/Jun/17

yes. this is the definition of W function.

$$\mathrm{yes}.\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of}\:\mathrm{W}\:\mathrm{function}. \\ $$

Commented by Tinkutara last updated on 12/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Commented by tawa tawa last updated on 12/Jun/17

logx^6 = (x/6) 6logx = x/6 36logx = x x^(36) = e^x 1 = x^(36) e^(−x) log .... Sir, i have been thinking. i have a question. What is my aim in solving using lambert function. Please give example as the number 1, i will use it to solve number 2. my problem here is that i dont even have starting point. Thank you sir. I want to learn it. The starting point is my problem. please do number 1 as example.

$$\mathrm{logx}^{\mathrm{6}} \:=\:\frac{\mathrm{x}}{\mathrm{6}} \\ $$$$\mathrm{6logx}\:=\:\mathrm{x}/\mathrm{6} \\ $$$$\mathrm{36logx}\:=\:\mathrm{x} \\ $$$$\mathrm{x}^{\mathrm{36}} \:=\:\mathrm{e}^{\mathrm{x}} \\ $$$$\mathrm{1}\:=\:\mathrm{x}^{\mathrm{36}} \:\mathrm{e}^{−\mathrm{x}} \\ $$$$\mathrm{log}\:.... \\ $$$$\mathrm{Sir},\:\mathrm{i}\:\mathrm{have}\:\mathrm{been}\:\mathrm{thinking}. \\ $$$$\mathrm{i}\:\mathrm{have}\:\mathrm{a}\:\mathrm{question}.\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{my}\:\mathrm{aim}\:\mathrm{in}\:\mathrm{solving}\:\mathrm{using}\:\mathrm{lambert}\:\mathrm{function}. \\ $$$$\mathrm{Please}\:\mathrm{give}\:\mathrm{example}\:\mathrm{as}\:\mathrm{the}\:\mathrm{number}\:\mathrm{1},\:\mathrm{i}\:\mathrm{will}\:\mathrm{use}\:\mathrm{it}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{number}\:\mathrm{2}. \\ $$$$\mathrm{my}\:\mathrm{problem}\:\mathrm{here}\:\mathrm{is}\:\mathrm{that}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{even}\:\mathrm{have}\:\mathrm{starting}\:\mathrm{point}. \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{it}. \\ $$$$\mathrm{The}\:\mathrm{starting}\:\mathrm{point}\:\mathrm{is}\:\mathrm{my}\:\mathrm{problem}.\: \\ $$$$\mathrm{please}\:\mathrm{do}\:\mathrm{number}\:\mathrm{1}\:\mathrm{as}\:\mathrm{example}. \\ $$

Commented by tawa tawa last updated on 12/Jun/17

Please give me one question on this, let me solve it.

$$\mathrm{Please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{one}\:\mathrm{question}\:\mathrm{on}\:\mathrm{this},\:\mathrm{let}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{it}. \\ $$

Commented by mrW1 last updated on 12/Jun/17

Solve x (1) log (x^6 )=(x/6) (2) log (x^3 )=(x/3)

$$\mathrm{Solve}\:\mathrm{x} \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{log}\:\left(\mathrm{x}^{\mathrm{6}} \right)=\frac{\mathrm{x}}{\mathrm{6}} \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{log}\:\left(\mathrm{x}^{\mathrm{3}} \right)=\frac{\mathrm{x}}{\mathrm{3}} \\ $$

Commented by mrW1 last updated on 12/Jun/17

To be able to use Lambert function we must re−arange the original equation into following form: A×e^A =B wherein A is a expression with x and B is a expression with constants only. The solution is then A=W(B) Example 1: log x^6 =(x/6) ⇒6log (±x)=(x/6) (± because x can be + or −) we consider +x at first ⇒log (x)=(x/(36)) ⇒((ln x)/(ln 10))=(x/(36)) ⇒ln x=((ln 10)/(36))x ⇒x=e^((ln 10 x)/(36)) ⇒xe^(−((ln 10 x)/(36))) =1 ⇒(−((ln 10 x)/(36)))e^(−((ln 10 x)/(36)) ) =−((ln 10)/(36)) this is the form A×e^A =B we need. the solution is −((ln 10 x)/(36))=W(−((ln 10)/(36))) ⇒x=−((36)/(ln 10))×W(−((ln 10)/(36))) we will get here 2 values for x, since W(−((ln 10)/(36))) has 2 values. similarly for the case −x we have ⇒log (−x)=(x/(36)) ⇒((ln (−x))/(ln 10))=(x/(36)) ⇒ln (−x)=((ln 10)/(36))x ⇒−x=e^((ln 10 x)/(36)) ⇒−xe^(−((ln 10 x)/(36))) =1 ⇒(−((ln 10 x)/(36)))e^(−((ln 10 x)/(36)) ) =((ln 10)/(36)) ⇒−((ln 10 x)/(36))=W(((ln 10)/(36))) ⇒x=−((36)/(ln 10))×W(((ln 10)/(36))) we will get here only one value for x. All solutions are thus x= { ((−((36)/(ln 10))×W(−((ln 10)/(36))) → 2 values)),((−((36)/(ln 10))×W(((ln 10)/(36))) → 1 value)) :}

$$\mathrm{To}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Lambert}\:\mathrm{function} \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{re}−\mathrm{arange}\:\mathrm{the}\:\mathrm{original} \\ $$$$\mathrm{equation}\:\mathrm{into}\:\mathrm{following}\:\mathrm{form}: \\ $$$$\mathrm{A}×\mathrm{e}^{\mathrm{A}} =\mathrm{B} \\ $$$$\mathrm{wherein}\:\mathrm{A}\:\mathrm{is}\:\mathrm{a}\:\mathrm{expression}\:\mathrm{with}\:\mathrm{x}\:\mathrm{and} \\ $$$$\mathrm{B}\:\mathrm{is}\:\mathrm{a}\:\mathrm{expression}\:\mathrm{with}\:\mathrm{constants}\:\mathrm{only}. \\ $$$$\mathrm{The}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{then} \\ $$$$\mathrm{A}=\boldsymbol{\mathrm{W}}\left(\mathrm{B}\right) \\ $$$$\mathrm{Example}\:\mathrm{1}:\:\mathrm{log}\:\mathrm{x}^{\mathrm{6}} =\frac{\mathrm{x}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{6log}\:\left(\pm\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\left(\pm\:\mathrm{because}\:\mathrm{x}\:\mathrm{can}\:\mathrm{be}\:+\:\mathrm{or}\:−\right) \\ $$$$\mathrm{we}\:\mathrm{consider}\:+\mathrm{x}\:\mathrm{at}\:\mathrm{first} \\ $$$$\Rightarrow\mathrm{log}\:\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{36}} \\ $$$$\Rightarrow\frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{ln}\:\mathrm{10}}=\frac{\mathrm{x}}{\mathrm{36}} \\ $$$$\Rightarrow\mathrm{ln}\:\mathrm{x}=\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\mathrm{x} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{e}^{\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}} \\ $$$$\Rightarrow\mathrm{xe}^{−\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}} =\mathrm{1} \\ $$$$\Rightarrow\left(−\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}\right)\mathrm{e}^{−\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}\:} =−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{form}\:\mathrm{A}×\mathrm{e}^{\mathrm{A}} =\mathrm{B}\:\mathrm{we}\:\mathrm{need}. \\ $$$$\mathrm{the}\:\mathrm{solution}\:\mathrm{is} \\ $$$$−\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}=\boldsymbol{\mathrm{W}}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\right) \\ $$$$\Rightarrow\mathrm{x}=−\frac{\mathrm{36}}{\mathrm{ln}\:\mathrm{10}}×\boldsymbol{\mathrm{W}}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\right) \\ $$$$\mathrm{we}\:\mathrm{will}\:\mathrm{get}\:\mathrm{here}\:\mathrm{2}\:\mathrm{values}\:\mathrm{for}\:\mathrm{x},\:\mathrm{since} \\ $$$$\boldsymbol{\mathrm{W}}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\right)\:\mathrm{has}\:\mathrm{2}\:\mathrm{values}. \\ $$$$ \\ $$$$\mathrm{similarly}\:\mathrm{for}\:\mathrm{the}\:\mathrm{case}\:−\mathrm{x}\:\mathrm{we}\:\mathrm{have} \\ $$$$\Rightarrow\mathrm{log}\:\left(−\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{36}} \\ $$$$\Rightarrow\frac{\mathrm{ln}\:\left(−\mathrm{x}\right)}{\mathrm{ln}\:\mathrm{10}}=\frac{\mathrm{x}}{\mathrm{36}} \\ $$$$\Rightarrow\mathrm{ln}\:\left(−\mathrm{x}\right)=\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\mathrm{x} \\ $$$$\Rightarrow−\mathrm{x}=\mathrm{e}^{\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}} \\ $$$$\Rightarrow−\mathrm{xe}^{−\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}} =\mathrm{1} \\ $$$$\Rightarrow\left(−\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}\right)\mathrm{e}^{−\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}\:} =\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}} \\ $$$$\Rightarrow−\frac{\mathrm{ln}\:\mathrm{10}\:\mathrm{x}}{\mathrm{36}}=\boldsymbol{\mathrm{W}}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\right) \\ $$$$\Rightarrow\mathrm{x}=−\frac{\mathrm{36}}{\mathrm{ln}\:\mathrm{10}}×\boldsymbol{\mathrm{W}}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\right) \\ $$$$\mathrm{we}\:\mathrm{will}\:\mathrm{get}\:\mathrm{here}\:\mathrm{only}\:\mathrm{one}\:\mathrm{value}\:\mathrm{for}\:\mathrm{x}. \\ $$$$ \\ $$$$\mathrm{All}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{thus} \\ $$$$\mathrm{x}=\begin{cases}{−\frac{\mathrm{36}}{\mathrm{ln}\:\mathrm{10}}×\boldsymbol{\mathrm{W}}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\right)\:\rightarrow\:\mathrm{2}\:\mathrm{values}}\\{−\frac{\mathrm{36}}{\mathrm{ln}\:\mathrm{10}}×\boldsymbol{\mathrm{W}}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\right)\:\rightarrow\:\mathrm{1}\:\mathrm{value}}\end{cases} \\ $$

Commented by Tinkutara last updated on 12/Jun/17

In the last line how can we determine that W(((−ln 10)/(36))) has 2 values while W(((ln 10)/(36))) has one value? Can it be without Geogebra?

$$\mathrm{In}\:\mathrm{the}\:\mathrm{last}\:\mathrm{line}\:\mathrm{how}\:\mathrm{can}\:\mathrm{we}\:\mathrm{determine} \\ $$$$\mathrm{that}\:{W}\left(\frac{−\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\right)\:\mathrm{has}\:\mathrm{2}\:\mathrm{values}\:\mathrm{while} \\ $$$${W}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{36}}\right)\:\mathrm{has}\:\mathrm{one}\:\mathrm{value}?\:\mathrm{Can}\:\mathrm{it}\:\mathrm{be} \\ $$$$\mathrm{without}\:\mathrm{Geogebra}? \\ $$

Commented by mrW1 last updated on 12/Jun/17

W(x) is the inverse function of f(x)=xe^x W(x) has only one value if x≥0 and 2 values if −(1/e)<x<0.

$$\mathrm{W}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{function}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{xe}^{\mathrm{x}} \\ $$$$\mathrm{W}\left(\mathrm{x}\right)\:\mathrm{has}\:\mathrm{only}\:\mathrm{one}\:\mathrm{value}\:\mathrm{if}\:\mathrm{x}\geqslant\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{2}\:\mathrm{values}\:\mathrm{if}\:−\frac{\mathrm{1}}{\mathrm{e}}<\mathrm{x}<\mathrm{0}. \\ $$

Commented by mrW1 last updated on 12/Jun/17

Commented by tawa tawa last updated on 12/Jun/17

2) ln(x^3 ) = (x/3) 3ln(±x) = (x/3) ln(±x) = (x/9) ((ln(±x))/(ln(10))) = (x/9) ln(±x) = ((xln(10))/9) ± x = e^((xln(10))/9) 1 = ±xe^(−((xln(10))/9)) −((xln(10))/9)e^(−((xln(10))/9)) = −((ln10)/9) −((xln(10))/9) = W[−((ln(10))/9)] x = − (9/(ln(10))) W[−((ln(10))/9)] x = − ((9(−0.3706))/(ln(10))) or x = ((9(−2.110))/(ln(10)))

$$\left.\mathrm{2}\right) \\ $$$$\mathrm{ln}\left(\mathrm{x}^{\mathrm{3}} \right)\:=\:\frac{\mathrm{x}}{\mathrm{3}} \\ $$$$\mathrm{3ln}\left(\pm\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\mathrm{3}} \\ $$$$\mathrm{ln}\left(\pm\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\mathrm{9}} \\ $$$$\frac{\mathrm{ln}\left(\pm\mathrm{x}\right)}{\mathrm{ln}\left(\mathrm{10}\right)}\:=\:\frac{\mathrm{x}}{\mathrm{9}} \\ $$$$\mathrm{ln}\left(\pm\mathrm{x}\right)\:=\:\frac{\mathrm{xln}\left(\mathrm{10}\right)}{\mathrm{9}} \\ $$$$\pm\:\mathrm{x}\:=\:\mathrm{e}^{\frac{\mathrm{xln}\left(\mathrm{10}\right)}{\mathrm{9}}} \\ $$$$\mathrm{1}\:=\:\pm\mathrm{xe}^{−\frac{\mathrm{xln}\left(\mathrm{10}\right)}{\mathrm{9}}} \\ $$$$−\frac{\mathrm{xln}\left(\mathrm{10}\right)}{\mathrm{9}}\mathrm{e}^{−\frac{\mathrm{xln}\left(\mathrm{10}\right)}{\mathrm{9}}} \:=\:−\frac{\mathrm{ln10}}{\mathrm{9}} \\ $$$$\:−\frac{\mathrm{xln}\left(\mathrm{10}\right)}{\mathrm{9}}\:=\:\mathrm{W}\left[−\frac{\mathrm{ln}\left(\mathrm{10}\right)}{\mathrm{9}}\right] \\ $$$$\:\mathrm{x}\:\:=\:−\:\frac{\mathrm{9}}{\mathrm{ln}\left(\mathrm{10}\right)}\:\mathrm{W}\left[−\frac{\mathrm{ln}\left(\mathrm{10}\right)}{\mathrm{9}}\right] \\ $$$$\mathrm{x}\:=\:\:−\:\frac{\mathrm{9}\left(−\mathrm{0}.\mathrm{3706}\right)}{\mathrm{ln}\left(\mathrm{10}\right)}\:\mathrm{or}\:\mathrm{x}\:=\:\:\frac{\mathrm{9}\left(−\mathrm{2}.\mathrm{110}\right)}{\mathrm{ln}\left(\mathrm{10}\right)}\: \\ $$

Commented by tawa tawa last updated on 12/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by mrW1 last updated on 12/Jun/17

in question (1) in log x^6 x can be + or −, thus log x^6 =6log (±x) we have totally 3 solutions. but in question (2) in log x^3 x must be +, thus log x^3 =3log x we have totally 2 solutions: x=−(9/(ln 10))×W(−((ln 10)/9))

$$\mathrm{in}\:\mathrm{question}\:\left(\mathrm{1}\right)\:\mathrm{in}\:\mathrm{log}\:\mathrm{x}^{\mathrm{6}} \:\mathrm{x}\:\mathrm{can}\:\mathrm{be}\:+\:\mathrm{or}\:−, \\ $$$$\mathrm{thus}\:\mathrm{log}\:\mathrm{x}^{\mathrm{6}} =\mathrm{6log}\:\left(\pm\mathrm{x}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{totally}\:\mathrm{3}\:\mathrm{solutions}. \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{in}\:\mathrm{question}\:\left(\mathrm{2}\right)\:\mathrm{in}\:\mathrm{log}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{x}\:\mathrm{must}\:\mathrm{be}\:+, \\ $$$$\mathrm{thus}\:\mathrm{log}\:\mathrm{x}^{\mathrm{3}} =\mathrm{3log}\:\mathrm{x} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{totally}\:\mathrm{2}\:\mathrm{solutions}: \\ $$$$\mathrm{x}=−\frac{\mathrm{9}}{\mathrm{ln}\:\mathrm{10}}×\mathrm{W}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{9}}\right) \\ $$

Commented by tawa tawa last updated on 12/Jun/17

Thank you sir. i understand.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{understand}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17

thank you so much dear mrW1. it is very beautiful and useful. god blees you master.

$${thank}\:{you}\:{so}\:{much}\:{dear}\:{mrW}\mathrm{1}. \\ $$$${it}\:{is}\:{very}\:{beautiful}\:{and}\:{useful}. \\ $$$${god}\:{blees}\:{you}\:{master}. \\ $$

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